The radius of a circle increases from 2 meters to 2.1 meters. Use differentials to estimate the increase in the area of this circle
I dont understand how to do this type of problem could someone help?
Area=PI*r^2
dArea=PI*2r*dr
so if dr is .1m
dA=PI(.1)(2*2)= .4PI estimate
check, actual change in area.
dA= Af-Ai=PI(2^2-2.1^2) = .41PI
huh? i dont get it
Certainly! To estimate the increase in the area of the circle using differentials, we can start by finding the formula for the area of a circle. The formula for the area of a circle is given by A = πr^2, where A is the area and r is the radius.
To find the increase in the area, we need to find the change in the area (∆A) when the radius changes from 2 meters to 2.1 meters.
We can start by finding the differential of the area (∆A). The differential of the area is given by dA = 2πr dr, where dA is the differential of the area, r is the radius, and dr is the change in the radius.
In this case, the initial radius (r) is 2 meters, and the change in the radius (dr) is 2.1 meters minus 2 meters, which is 0.1 meters.
We can now substitute these values into the differential of the area equation:
dA = 2πr dr
dA = 2π(2) (0.1)
dA = 0.4π
So, the differential of the area (∆A) is approximately 0.4π.
To estimate the actual increase in the area, we can substitute the initial radius (r) of 2 meters into the area formula and calculate the initial area (A1), and then substitute the new radius (r+dr) of 2.1 meters into the area formula and calculate the new area (A2). Finally, we subtract the initial area from the new area to find the actual increase in the area (∆A).
Using the area formula A = πr^2:
A1 = π(2)^2
A1 = 4π (since π(2)^2 is the same as 4π)
A2 = π(2.1)^2
A2 = 4.41π (since π(2.1)^2 is the same as 4.41π)
∆A = A2 - A1
∆A = 4.41π - 4π
∆A ≈ 0.41π
Therefore, the actual increase in the area of the circle is approximately 0.41π (or about 1.29 square meters).